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Measures and alternating multilinear forms

  1. Dec 18, 2008 #1

    mma

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    On an n-dimensional vector space an alternating n-form defines a measure. However a measure can be defined on its own right, without mentioning any alternating form. My question is that what condition must a measure satsfy that it can be originated from an alternating multilinear form. I mean an analogue that of a norm can be originated from a symmetric bilinear form if and only if it satisfies the parallelogram identity.
     
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  3. Dec 20, 2008 #2

    mma

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    I put the question the other way around.

    We learn in the school that the area of a polygon is determined by the length of its edges and its shape, i.e. the angles between the edges. Length and angles are determinded by the scalar product. So we learn the notion of area based on the scalar product.

    However, area (or at least the ratio of areas) is determined solely by the linear structure of the vector space.

    My queston is now, how coul'd we introduce in a natural way the notion or area (or volume, etc.) without using the notion of norm or angles or scalar product ?

    Of course, the emphasis is on the word natural, because otherwise we could simply say that area is an alternating bilinear form.
     
  4. Dec 20, 2008 #3

    Hurkyl

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    I think the Haar measure is what you're looking for.
     
  5. Dec 20, 2008 #4

    mma

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    Do you mean that we regard our vector space as the group of translations, and we apply the existence and uniquiness of the Haar-measure on this group?

    This is interesting. The requirement of the translation invariance alone determines the area or volume function up to multiplication by a positive constant?
     
  6. Dec 20, 2008 #5

    Hurkyl

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    That's what it looks like.

    Oh, and don't forget the regularity conditions! Although (except for the finiteness condition) I don't know what can go badly if you omit them.
     
  7. Dec 20, 2008 #6

    mma

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    Yes, of course. These are the "natural" features of a measure, independently of any group or vector space strucrure. And if we add the translation invariance, then the ratio of the areas of two parallelograms is determined unambiguously. Super! Could you show a relatively simple proof for this?
     
  8. Dec 20, 2008 #7

    Hurkyl

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    It depends on what you mean by simple! I haven't pushed it through to completion, but my best lead on a 'simple' proof would be to pick a basis, which thus lets you define a notion of an n-parallelepiped. If you have two translation-invariant, regular measures, then I believe you can show that for any n-parallelepiped whose faces are parallel to the fundamental one, the ratio of the measures is a fixed constant. I presume you can push that through to show their ratio on any measurable set is a fixed constant.
     
  9. Dec 21, 2008 #8

    gel

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    the only regularity assumption you need is that the measure is finite and nonzero on a given bounded and and nonempty open set. Or, just sigma-finite is enough.

    For the proof, suppose [itex]\mu[/itex] and [itex]\nu[/tex] are two such measures.

    [tex]
    \begin{align*}
    \mu(A)\nu(B) &= \int\int 1_{\{x\in A,y\in B\}}d\mu(x)d\nu(y)\\
    &= \int\int 1_{\{x-y\in A,y\in B\}}d\mu(x)d\nu(y)\ \ (x \rightarrow x - y)\\
    &= \int\int 1_{\{-y\in A,y+x\in B\}}d\nu(y)d\mu(x)\ \ (y\rightarrow y + x)\\
    &= \int\int 1_{\{-y\in A,x\in B\}}d\mu(x)d\nu(y)\ \ (x\rightarrow x-y)\\
    &= \nu(-A)\mu(B)
    \end{align*}
    [/tex]

    so,
    [tex]
    \mu(B)/\nu(B) = \mu(A)/\nu(-A)
    [/tex]
    is independent of the set B chosen, and [itex]\mu[/itex] is proportional to [itex]\nu[/itex].
     
    Last edited: Dec 21, 2008
  10. Dec 21, 2008 #9

    Hurkyl

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    Ha! That's (roughly) the swap-in-place trick from computer lore. It's really neat to see it useful elsewhere!
     
  11. Dec 21, 2008 #10

    mma

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    Very tricky! But we arrived to [tex]\mu(B)/\nu(B) = \mu(A)/\nu(-A)[/tex] and not to [tex]\mu(B)/\nu(B) = \mu(A)/\nu(A)[/tex]. So we need additionally suppose the [tex]\nu(A)=\nu(-A)[/tex] mirroring invariance too. Or can this also be proved from our original conditions?
     
  12. Dec 21, 2008 #11

    gel

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    Fix any set A with [itex]\nu(-A)\not=0[/itex] and set [itex]\lambda=\mu(A)/\nu(-A)[/itex]. Then, my proof showed that [itex]\mu(B)=\lambda\nu(B)[/itex] for every set B, so [itex]\mu,\nu[/itex] are proportional.

    So, no you don't need to assume `mirror invariance', but you can prove it with a couple of extra lines (try putting [itex]\mu=\nu,A=B[/itex] into my equation).
     
    Last edited: Dec 21, 2008
  13. Dec 21, 2008 #12

    mma

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    Yes, it was evident, sorry for the stupid question.
    Thank you Gel and Hurkyl!
     
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